= Solution
The <complex velocity> obtained from part (b) is
$$
w'(z)=-\frac A{(z-b)^2}+\frac B{(z-c)^2}.
$$
For a nonzero <planar potential doublet>, $B/A=(a^2/b^2)e^{-2i\alpha}$. Its zeros therefore satisfy
$$
\frac{z-c}{z-b}=t_\pm,\qquad t_\pm=\pm\frac ab e^{-i\alpha},
$$
and hence
$$
\boxed{z_\pm=\frac{c-bt_\pm}{1-t_\pm}=a\frac{a/b\mp e^{-i\alpha}}{1\mp(a/b)e^{-i\alpha}}.}
$$
The denominators are nonzero because $a/b<1$. For either sign,
$$
|a/b\mp e^{-i\alpha}|^2=|1\mp(a/b)e^{-i\alpha}|^2,
$$
so $|z_\pm|=a$. They are distinct because $t_+\neq t_-$ and $t\mapsto(c-bt)/(1-t)$ is <injective>, its <determinant> being $c-b\neq0$. Clearing the velocity denominators gives a quadratic with these two <roots>, so there are no further finite <stagnation points>. Thus \b[both <stagnation points> lie on the obstacle <boundary>].
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